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Howdy suppose I have the system

$\rho u_t + p_x=0 \\ \rho v_t + p_y=0 \\ p_t + K(u_x+v_y)=0$

I'm interesting in showing that

$u_{tt} = \frac{K}{\rho}(u_{xx} + u_{yy}) \\ v_{tt} = \frac{K}{\rho}(v_{xx} + v_{yy})$

So far what I have concluded is that

$u_{tt} = \frac{K}{\rho}(u_{xx}+v_{yx})\\ v_{tt} = \frac{K}{\rho}(u_{xy}+v_{yy})$

So really to be complete I just have to show that

$v_{yx}=u_{yy} \\ u_{xy}=v_{xx}$

Now my question. I've shown that

$\frac{\partial}{\partial t }(u_y) = \frac{\partial}{\partial t }(v_x)$

My problem is that this doesn't mean in my mind that

$u_y=v_x$

all it shows is that

$u_y=v_x +c(x,y)$

Can someone help me make an argument that $c(x,y)=0$ or an approach that I'm missing?

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Let's differentiate the last equation $p_t + K(u_x + u_y) = 0$ with respect to $t$: \begin{aligned} p_{tt} + K(u_{xt} + v_{yt}) &= 0 \, . \end{aligned} Now, we differentiate the two first equations with respect to $x$ and $y$, respectively: \begin{aligned} \rho u_{tx} + p_{xx} &= 0 \, ,\\ \rho v_{ty} + p_{yy} &= 0 \, . \end{aligned} Using the equality of mixed derivatives, one obtains $$ p_{tt} - \frac{K}{\rho}(p_{xx} + p_{yy}) = 0 \, . $$ However, differentiating $p_t + K(u_x + u_y) = 0$ w.r.t. $x$ and w.r.t. $y$, as well as the two first equations w.r.t. $t$, one obtains \begin{aligned} u_{tt} - \frac{K}{\rho}(u_{xx} + v_{yx}) &= 0 \, ,\\ v_{tt} - \frac{K}{\rho}(u_{xy} + v_{yy}) &= 0 \, . \end{aligned}

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